Answer:
M.Mass = 3.66 g/mol
Data Given:
M.Mass = M = ??
Density = d = 0.1633 g/L
Temperature = T = 273.15 K (Standard)
Pressure = P = 1 atm (standard)
Solution:
Let us suppose that the gas is an ideal gas. Therefore, we will apply Ideal Gas equation i.e.
P V = n R T ---- (1)
Also, we know that;
Moles = n = mass / M.Mass
Or, n = m / M
Substituting n in Eq. 1.
P V = m/M R T --- (2)
Rearranging Eq.2 i.e.
P M = m/V R T --- (3)
As,
Mass / Volume = m/V = Density = d
So, Eq. 3 can be written as,
P M = d R T
Solving for M.Mass i.e.
M = d R T / P
Putting values,
M = 0.1633 g/L × 0.08205 L.atm.K⁻¹.mol⁻¹ × 273.15 K / 1 atm
M = 3.66 g/mol
Answer:
living organisms whose genetic material has been artificially manipulated in a laboratory through genetic engineering
Explanation:
Answer:
108.43 grams KNO₃
Explanation:
To solve this problem we use the formula:
Where
- ΔT is the temperature difference (14.5 K)
- Kf is the cryoscopic constant (1.86 K·m⁻¹)
- b is the molality of the solution (moles KNO₃ per kg of water)
- and<em> i</em> is the van't Hoff factor (2 for KNO₃)
We <u>solve for b</u>:
- 14.5 K = 1.86 K·m⁻¹ * b * 2
Using the given volume of water and its density (aprx. 1 g/mL) we <u>calculate the necessary moles of KNO₃</u>:
- 275 mL water ≅ 275 g water
- moles KNO₃ = molality * kg water = 3.90 * 0.275
- moles KNO₃ = 1.0725 moles KNO₃
Finally we <u>convert KNO₃ moles to grams</u>, using its molecular weight:
- 1.0725 moles KNO₃ * 101.103 g/mol = 108.43 grams KNO₃
The empirical formula : MnO₂.
<h3>Further explanation</h3>
Given
632mg of manganese(Mn) = 0.632 g
368mg of oxygen(O) = 0.368 g
M Mn = 55
M O = 16
Required
The empirical formula
Solution
You didn't include the pictures, but the steps for finding the empirical formula are generally the same
- Find mol(mass : atomic mass)
Mn : 0.632 : 55 = 0.0115
O : 0.368 : 16 =0.023
- Divide by the smallest mol(Mn=0.0115)
Mn : O =
The empirical formula : MnO₂
The answer is [Ne] 3s^2 3p^5 because chlorine is the fifth element in the 3rd row of elements in in p orbital
